Speed and velocity refer to the motion of an object. Distance-time and velocity-time graphs can be a useful way of analysing motion.

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If an object moves along a straight line, its motion can be represented by a velocity-time graph. The gradient of the line is equal to the acceleration of the object.

The table shows what each section of the graph represents:

Section of graph | Gradient | Velocity | Acceleration |
---|---|---|---|

A | postive | increasing | postive |

B | zero | constant | zero |

C | negative | decreasing | negative |

D | zero | stationary (at rest) | zero |

Describe the motion of the vehicle in the graph at the three stages of its journey.

Between 0 s and 4 s the vehicle is accelerating.

This means that:

Between 4 s and 7 s the vehicle has a constant velocity of 8 m/s.

Between 7 s and 10 s, the vehicle is decelerating so:

The displacement of an object can be calculated from the area under a velocity-time graph.

The area under the graph can be calculated by:

- using geometry (if the lines are straight)
- counting the squares beneath the line (particularly if the lines are curved)

Calculate the total displacement of the object whose motion is represented by the velocity-time graph below.

The displacement can be found by calculating the total area of the shaded sections below the line.

Find the area of the triangle:

Find the area of the rectangle:

base × height

(10 - 4) × 8 = 48 m^{2}

Add the areas together to find the total displacement:

(16 + 48) = 64 m